MOMENT DISTRIBUTION BY Hardy Cross, 1930 DISPLACEMENT MEHTOD OF ANALYSIS

Sign Convention: Clockwise moment: Positive Counterclockwise moment: Negative Fixed-End Moments (FEMS): MAB +MBA

Far End Fixed Member Stiffness Factor:

Far End Hinged  A M M’

DISTRIBUTION FACTOR (DF) M M1 M2 M3 K1 K2 K3

Determine the moment at each end 10 kN 2 kN/m 1.5 kN/m 3.75 m 3.75 m 5 m 6.25 m EI = constant

10 kN 2 kN/m 1.5 kN/m 3.75 m 3.75 m 5 m 6.25 m

10 kN 2 kN/m 1.5 kN/m 3.75 m 3.75 m 5 m 6.25 m

10 kN 2 kN/m 1.5 kN/m 3.75 m 3.75 m 5 m 6.25 m

10 kN 2 kN/m 1.5 kN/m 3.75 m 3.75 m 5 m 6.25 m

10 kN 2 kN/m 1.5 kN/m AB 3.75 m 3.75 m 5 m 6.25 m SPAN BC CD L 7.5 m 1/7.5 1/5 1/6.25 JOINT A B C D DF 0.4 0.6 0.556 0.444 1 FEM -9.375 9.375 -4.167 4.167 -4.883 4.883 BAL CO 5.208 -0.716 -2.083 -3.125 0.398 0.318 -4.883 -1.042 0.199 -1.562 -2.442 0.159 0.199 -4.004 -0.080 -0.119 2.226 1.778 -0.159 -0.040 1.113 -0.06 -0.080 0.889

10 kN 2 kN/m 1.5 kN/m 3.75 m 3.75 m 5 m 6.25 m JOINT A B C D DF 0.4 0.4 0.6 0.556 0.444 1 CO -0.040 1.113 -0.06 -0.08 0.889 -0.140 -0.889 BAL -0.445 -0.668 0.078 0.062 -0.223 0.039 -0.334 0.031 -0.779 -0.016 -0.023 0.433 0.346 -0.031 -0.008 0.217 -0.012 0.173 -0.028 -0.087 -0.130 0.016 0.012 -0.173 -0.044 0.008 -0.065 0.006

10 kN 2 kN/m 1.5 kN/m 3.75 m 3.75 m 5 m 6.25 m JOINT A B C D DF 0.4 0.4 0.6 0.556 0.444 1 CO -0.045 0.008 -0.065 -0.087 0.006 0.152 -0.006 BAL -0.003 -0.005 0.085 0.067 -0.002 0.043 0.034 -0.017 -0.026 0.003 -0.034 SUM -10.753 6.642 -6.642 5.37 -5.391 MOM -10.75 6.60 -6.60 5.40 -5.40

MODIFIED STIFFNESS METHOD 10 kN 2 kN/m 1.5 kN/m 3.75 m 3.75 m 5 m 6.25 m SPAN AB BC CD L 7.5 m 5 m 6.25m K 1/7.5 1/5=0.2 (3/4)1/6.25=0.12 JOINT A B C D DF 0.4 0.6 0.625 0.375 1 FEM -9.375 9.375 -4.167 4.167 -4.883 4.883 BAL CO 5.208 -0.716 -2.083 -3.125 0.448 0.269 -4.883 -1.042 0.224 -1.563 -2.442 0.224 -4.005 -0.09 -0.134 2.503 1.502 -0.045 1.252 -0.067 PRESENTED BY: Engr. Carol E. Dungca

AB SPAN BC CD JOINT A B C D DF 0.4 0.6 0.625 0.375 1 CO -0.45 1.252 0.4 0.6 0.625 0.375 1 CO -0.45 1.252 -0.067 BAL -0.501 -0.751 0.042 0.025 -0.251 0.021 -0.376 -0.008 -0.013 0.235 0.141 -0.004 0.118 -0.007 -0.047 -0.071 0.004 0.003 SUM -10.717 6.646 -6.646 5.386 -5.386 FEM

AB SPAN BC CD L 6 m 9 m 4m K 1/6 (3/4)(1/9)=1/12 JOINT A B C D DF 2/3 10kN/m 30 kN A 6m B 9m C 4m D SPAN AB BC CD L 6 m 9 m 4m K 1/6 (3/4)(1/9)=1/12 JOINT A B C D DF 2/3 1/3 1 FEM -67.5 67.5 -120 BAL CO SUM -67.5 -52.5 45 22.5 52.5 22.5 26.25 26.25 -17.5 -8.75 -8.75 13.75 27.5 -27.5 120 -120 13.75 27.5 -27.5 120 -120

Determine the member end moment for the three span continuous beam shown due to the uniformly distributed load and due to the support settlement of 15 mm at B, 36 mm at C, and 18 mm at D. E = 200 GOPa, I = 1705 x 106 mm4 32 kN/m A 5m B 5m C 5m D

32 kN/m 0.015 m 0.036 m 0.018 m 0.021 m 32 kN/m ∆ ∆

∆ ∆ PRESENTED BY: Engr. Carol E. Dungca

∆ ∆ PRESENTED BY: Engr. Carol E. Dungca

SPAN AB BC CD K ¾(1/5) = 0.15 1/5 = 0.2 ¾(1/5) JOINT A B C D DF 1 0.429 0.571 FEM(LOAD) FEM(SETTLE) -66.67 -1227.6 66.67 -1718.64 1473.12 FEM(total) -1294.27 -1160.93 -1785.31 -1651.97 1406.45 1539.79 BAL CO -2946.24 -245.52 1294.27 1263.937 1682.303 140.192 105.328 -1539.79 647.135 70.096 841.152 -769.895 717.231 71.257 -307.692 -409.539 -40.688 -30.569 -20.344 -204.770 -20.344 -204.770 8.728 11.616 116.924 87.846 58.462 5.808 58.462 5.808 -25.080 -33.382 -3.316 -2.492 PRESENTED BY: Engr. Carol E. Dungca

SPAN AB BC CD K ¾(1/5) = 0.15 1/5 = 0.2 ¾(1/5) JOINT A B C D DF 1 0.429 0.571 BAL CO -1.658 -16.691 0.711 0.947 9.531 7.160 4.766 0.474 -2.045 -2.721 -0.271 -0.203 -0.136 -1.379 0.058 0.078 0.787 0.592 -25.080 -33.382 -3.316 -2.492 PRESENTED BY: Engr. Carol E. Dungca

SPAN AB BC CD K ¾(1/5) = 0.15 1/5 = 0.2 ¾(1/5) JOINT A B C D DF 1 0.429 0.571 BAL 0.058 0.078 0.787 0.592 CO 0.394 0.039 -0.169 -0.225 -0.022 -0.017 -0.011 -0.113 0.005 0.006 0.065 0.048 0.033 -0.014 -0.019 SUM/FEM 424.62 -424.644 -804.234 804.234 PRESENTED BY: Engr. Carol E. Dungca

10 kN 5kN/m 50 kN m JOINT A B C D MEMBER AB BA BC CB CD DC K 0.188 A 2EI B 8m C 8m D 4m 4m 3EI 3EI JOINT A B C D MEMBER AB BA BC CB CD DC K (3/4)(2/8)= 3/16= 0.188 0.188 0.375 0.281 DF 1 0.334 0.666 0.572 0.428 JT. COUPLE CO FEM -10 10 -26.667 26.667 BAL 50 16.7 33.300 16.650 -16.667 16.650 10 5.567 11.100 -9.524 -7.126 -26.667 5 -4.762 5.550 -13.334 PRESENTED BY: Engr. Carol E. Dungca

JOINT A B C D MEMBER AB BA BC CB CD DC DF 1 0.334 0.666 0.572 0.428 CO -4.762 5.550 -13.334 0.238 -7.784 BAL -0.079 -0.159 4.452 3.332 2.226 -0.080 -0.743 -1.483 0.046 0.034 0.023 -0.741 -0.008 -0.015 0.424 0.317 FEM 36.437 13.563 43.444 PRESENTED BY: Engr. Carol E. Dungca

Determine the internal moment acting at each joint Determine the internal moment acting at each joint. Assume A,D, and E are pinned and B and C are fixed joints. E = 29 x 103 ksi PRESENTED BY: Engr. Carol E. Dungca

JOINT A B C E D AB BA BD BC CB CE EC DB K 0.125 0.047 0.056 0.078 DF 1 MEMBER AB BA BD BC CB CE EC DB K 0.125 0.047 0.056 0.078 DF 1 0.548 0.206 0.246 0.418 0.582 FEM -12 12 -108 108 -30 30 BAL CO -96 78 12 52.608 19.776 23.616 -32.604 -45.396 -30 6 -15 -16.302 11.808 -10.302 -3.192 5.645 2.122 2.534 1.334 1.858 0.667 1.267 0.667 1.267 -0.366 -0.137 -0.164 -0.530 -0.737 75.887 21.761 -97.649 89.275 -89.275 PRESENTED BY: Engr. Carol E. Dungca